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For mass density $\rho$, the Newtonian gravitational potential satisfies the Poisson equation ${\nabla }^2\Phi =4\pi G\rho$. The gravitational acceleration is $-\nabla \Phi$. An isolated system also requires the exterior vacuum solution and decay at infinity; solving the interior equation alone does not fix a globally physical field.

A radial Newtonian gravitational potential $\Phi =-A{r}^{-\beta }$ produces inward acceleration $A\beta r^{-\beta -1}$. The point-mass case has $\beta =1$ and $A=GM$. Treating the power-law gravitational potential as prescribed does not require it to be sourced by the flowing gas.

In a fixed central Newtonian gravitational potential, pressureless free fall has speed $u=a{R}^{-1/2}$, and randomly moving stars have typical gas-relative speed $V=b{R}^{-1/2}$. A steady, undepleted spherical mass flux $A$ gives $\rho =A{R}^{-3/2}/(4\pi a)$. The Bondi--Hoyle--Lyttleton accretion rate therefore becomes $\dot{M}=\lambda M^2$ with $\lambda =G^{2}A/(a{b}^3)$, independent of radius. Integrating $d(1/M)/dt=-\lambda$ gives the displayed solution before $\lambda M_{0}t=1$. Its finite-time divergence signals failure of the fixed potential, undepleted gas supply and negligible feedback assumptions; it is not a physically unlimited mass supply.

For a point-mass companion at separation $a$, subtract the constant and linear terms of its Newtonian gravitational potential in a frame accelerating with the target's center. The leading residual isIt is degree two in spherical harmonics. In a circular orbit with angular frequency ${\omega }_o$, its time-dependent $m=\pm 2$ components force at $2{\omega }_o$ in a frame with no spin; there is also a static $m=0$ component. This is the leading approximation in $r/a$, and the tidal amplitude may be absorbed into a harmonic-dependent constant.

Let a test particle start at rest at radius $r_0$ outside a fixed spherically symmetric source. The spherical shell theorem gives the exterior Newtonian gravitational potential $\Phi (r)=-\mu /r$, where $\mu =GM$ is the source's standard gravitational parameter. Zero initial transverse velocity makes the motion radial. Conservation of mechanical energy per unit mass givesThe minus sign selects inward motion; the positive square root is its speed. For a fall from $r_0=2R$ to the surface $r=R$, that speed is $\sqrt{\mu /R}$. In the limit of release from rest infinitely far away, the familiar free-fall speed is $\sqrt{2\mu /r}$, equal in magnitude to the local escape velocity. A finite starting radius gives a smaller speed.

The arrival time follows from the same first integral. Set $r=r_0{\cos }^2\eta$. ThenFor arrival at $R>0$ this is finite, and the field formula is used only outside the source. For $r_0=2R$, $\eta =\pi /4$ givesKeeping the smaller initial gravitational magnitude $g(r_0)=\mu /(4R^2)$ constant would predict $\sqrt{8}\sqrt{R^3/\mu }$, about $10.0\%$ longer. The actual acceleration increases during the fall.

The curves compare a shallow fall, $h/R=0.02$, with a fall from $r_0=2R$, so $h/R=1$. Each comparison holds the release height and initial gravitational magnitude fixed. The constant-gravity model is close for the shallow fall and increasingly inaccurate for the larger one. This is also a practical use of nondimensionalization: the normalized trajectories depend on $h/R$, while $\sqrt{R^3/\mu }$ supplies the dimensional time scale.

For a central-force radial energy equation, choose a monotone radius $q=q(r)$ and clock $\tau$ such that $d\tau /dt=(q/r)dq/dr$. Then $q^{\prime }=(r/q)\dot{r}$, and multiplying the old energy equation by $r^2/q^2$ preserves the centrifugal form $L^2/(2q^2)$. A split of the remaining terms into constant energy and a new Newtonian gravitational potential produces a transformed radial orbit.

A regular solid harmonic is $r^l{Y}_l^m(\theta ,\varphi )$: it is homogeneous of degree $l$ and solves the Laplace equation. Thus $\mathbf{r}\cdot \nabla F=lF$ and ${\nabla }^{2}F=0$. These identities turn angular spherical harmonic expansions of displacement, density and Newtonian gravitational potential into radial equations.

For uniform rotation, the Newtonian gravitational potential plus the centrifugal potential generates the non-Coriolis part of the equation of motion in a rotating frame. Stationary particles obey $\nabla {\Phi }_{\mathrm{eff}}=0$. The Coriolis force changes perturbation dynamics, so curvature of this potential alone is not a general rotating-frame stability criterion.

Self-gravity is the gravitational acceleration produced by a system's own mass density. It is obtained by solving the Poisson equation for Newtonian gravity for that density, rather than treating the Newtonian gravitational potential as an externally fixed field. It competes with pressure and rotational support in gravitational instability of an astrophysical disk.

A smooth radial compressible flow with a fixed adiabatic exponent conserves $r^2\rho u$ and $p/{\rho }^{\gamma }$. In a prescribed Newtonian gravitational potential, its Bernoulli function is $u^2/2+c^2/(\gamma -1)+\Phi$, with $c$ the adiabatic sound speed. Eliminating the density derivative yields $(u-c^2/u)u^{\prime }=2c^2/r-{\Phi }^{\prime }$.

For a uniform-density sphere with incompressible flow of radius $R$ with one surface spherical harmonic displacement ${\xi }_r(R)=lA{R}^{l-1}{Y}_l^m$, the bulk density perturbation vanishes but the surface contributes $\delta \rho =\rho {\xi }_r(R)\delta (r-R)$. Continuity of the Newtonian gravitational potential and the Poisson equation give $[{\partial }_r\delta \Phi {]}_{\mathrm{out}-\mathrm{in}}=4\pi G\rho {\xi }_r(R)$. The regular interior and decaying exterior solutions areThe coefficient follows because the derivative jump is $-(2l+1)C{R}^{l-1}{Y}_l^m$. This surface source is why neglecting the bulk density perturbation does not justify neglecting the perturbed gravity.

The acceleration tidal tensor is $T_{jk}={\partial }_k{g}_j=-{\partial }_j{\partial }_k\Phi$ for gravitational acceleration $\mathbf{g}=-\nabla \Phi$. It describes differential acceleration $\Delta \mathbf{g}\simeq T\Delta \mathbf{r}$. Some authors instead call the positive Hessian matrix ${\partial }_j{\partial }_k\Phi$ the tidal tensor, reversing this sign convention. For a point mass and the positive potential convention $\mathbf{g}=\nabla \psi$, the tensor is $Gm(3\mathbf{R}{\mathbf{R}}^{\mathsf{T}}/R^5-1/R^3)$.

The tidal tensor is the Jacobian matrix of the Newtonian gravitational field, rather than the gravitational acceleration itself. Its entries have units ${\mathrm{s}}^{-2}$: multiplying by a displacement produces an acceleration. For a smooth Newtonian gravitational potential, the tidal tensor is a symmetric matrix, since mixed partial derivatives commute. A spatially uniform gravitational acceleration has zero tidal tensor, however large that acceleration is.

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.

A stationary axisymmetric galactic distribution function depending on the specific orbital energy and axial specific angular momentum is a two-integral distribution. These quantities are conserved in a time-independent axisymmetric Newtonian gravitational potential, so the Collisionless Boltzmann equation holds by the chain rule. Even dependence on $L_z$ removes net azimuthal streaming velocity; an odd part can supply rotation without changing the mass density.